SPH 304: Electrodynamics I

Credit Hours: 3
Prerequisites: SPH 102: Electricity and Magnetism I SPH 202: Electricity and Magnetism II.

COURSE PURPOSE
The purpose of this course is to equip student with the theoretical and experimental
electromagnetic theory concepts. The students will be introduced to the main ideas of
electrodynamics and be taught the basic principles of electrodynamics.
Expected learning outcomes
At the end of this course, students should be able to,
1. Identify and describe the fundamental concepts learnt in electromagnetic theory.
2. Solve a range of numerical problems involving the learnt concepts in Electromagnetic theory.
3. Perform experiments based on Electromagnetic theory, analyze experimental data and write
laboratory report.

COURSE CONTENT
1. Electric field vectors.
2. Conservation of charge and the continuity equation.
3. Polarizability and dielectric tensors
4. Magnetic field vectors; Lorentz force. .
5. Field energy in free space.
6. Poynting’s vector.
7. Energy density.
8. Laplace and Poisson equations and solutions in cylindrical and spherical coordinates.
9. Maxwell’s equations.
10. Piezoelectric and ferroelectric properties.
11. Paramagnetism, Diamagnetism and Ferromagnetism.

Mode of Delivery
The course will be delivered through Lectures, Tutorials, Practical demonstrations and Hands on
Laboratory sessions.

Mode of Assessment
Mode of assessment
Continuous Assessment 30 %
End of Semester examination 70 %
Total 100 %

Instructional Material and/or Equipment

White board and white board markers, Laboratory manuals, Laboratory Equipment and
Apparatus. (Name them)

Core References
Mansuripur M and Mashud (2011), Field, Force, Energy and Momentum in Classical
Electrodynamics. Bentharm Science Publishers.
Geyi Wen, (2010), Foundations of Applied Electrodynamics

Recommended References
David J. Griffiths (1999). Introduction to electrodynamics. Prentice Hall, Upper Saddle River,
New Jersey, ISBN 0-13-805326-X.
Jackson, John D. (1998). Classical Electrodynamics (3rd ed.). Wiley. ISBN 0-471-30932-X.
Purcell, Edward M. (1985). Electricity and Magnetism Berkeley Physics Course Volume 2 (2nd
ed.). McGraw-Hill. ISBN 0-07-004908-4.
Relevant Website;
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html.
WEEK ONE

1. Electric and Magnetic field vectors.

1.1 Electric field vectors.
1.1.1: Coulomb’s Law
Consider a test charge and a single point charge , which are at rest a distance apart
as shown in figure 1.

Figure 1: Separation of and .

 The force on due to is given by Coulomb’s Law;

̂ ………………………………………….(1)

 Coulomb’s law the force is proportional to the product of the charges and
inversely proportional to the square of the separation distance.
 The sign is a constant called the permittivity of free space and has the value,
.
 The force points along the path from to, . It is repulsive if and have the
same sign and attractive if they have opposite sign.
 ⃗ is the separation vector from , the position vector of , to the position
vector of as shown in figure 2.

Figure 2: Vector diagram for the charges and .

The separation is given by, ⃗ ………………………………………….(2)
Note that;
⃗ - is the separation vector. It’s the magnitude is given by,
⃗⃗⃗⃗⃗
| | | | √
̂ -is the unit vector in the direction of the separation distance. It is given by
⃗⃗⃗⃗⃗
̂
⃗⃗⃗⃗⃗
| | | | √

1.1.2: Electric Field of Discrete distribution of charges

Consider several point charges each at a distance of
respectively from a test charge as shown in figure 3
We want to determine the total force on due to all the point charges

Figure 3: Vector diagram for several point charges.

We apply the principle of superposition which states that the interaction between any two
charges is completely unaffected by the presence of other charges.
We separately calculate the forces, , and then take the vector sum i.e,

̂ ̂ ̂

⌈ ̂ ̂ ̂ ⌉
⃗

Where ⃗ ⌈ ̂ ̂ ̂ ⌉

∑ ̂

⃗ is the electric field of the source charges. It is a function of position since the
separation vectors ⃗ depend on the location of of the field point, . The electric field,
⃗ does not depend on the test charge
1.1.3: Electric Field of continuous distribution of charge
If the charge is distributed continuously over some region, the sum becomes an integral
i.e
⃗ ∫ ̂

The equation changes according to the region in question, i.e line, surface, or volume as
follows,
(i) Line charge
If the charge is spread out along a line i.e

Figure 4: Line charge distribution

The charge per unit length or charge density is denoted by lambda, and where
is an element of length along the line. Also
The electric field for line charge distribution is therefore given by,
⃗ ∫ ̂

Note that the unit vector ̂ is the vector from to the field point, P ( ̂ .

(ii) Surface charge

If the charge is smeared over a surface i.e

Figure 5: surface charge distribution

The charge per unit area or charge density is denoted by sigma, and where
is an element of the area on the surface. Also
The electric field for surface charge distribution is therefore given by,
⃗ ∫ ̂

Note that the unit vector ̂ is the vector from to the field point, P ( ̂ .
(iii) Volume charge
If the charge fills up a volume i.e

Figure 6: Volume charge distribution

The charge per unit volume or charge density is denoted by rho, and where
is an element of the volume. Also
The electric field for volume charge distribution is therefore given by,
⃗ ∫ ̂

Note that the unit vector ̂ is the vector from to the field point .
In all the cases the unit vector ̂ is not constant but it’s direction depends on the source
point and so cannot be taken outside the integrals.
Example 1
Two point charges and is are placed along the as shown in the diagram.
Calculate the electric field ⃗ at any point on the . What is ⃗ when ?

⃗ ∑ ̂

⃗⃗⃗
But, ̂ ⃗⃗⃗⃗⃗ | |
and | | √ hence,
| | √

⃗ ∑ ⃗

̂ ̂ ̂
⃗ * +

Collecting like terms together,

 ̂
⃗ *( ) ̂ +

when then ⃗ ̂

Example 2
A uniformly charged rod with linear charge density lies on the positive axis and
extends from the origin to a point . Calculate the electric field at any point on the
Z above the rod.
Solution

⃗ ∫

But, , ̂ and , therefore,

̂
⃗ ∫

Let such that, , hence,

̂
⃗ ∫

̂
+

̂
[ ]
WEEK TWO

Flux of the electric field ⃗ )-Gauss’s law

Consider a single point charge located at the origin. The electric field is;

⃗ ̂

The flux of ⃗ through a surface is given by,

∫⃗

Where is a small element of the surface and ⃗ is perpendicular to as shown in

the figure below.

Figure 7: Flux of ⃗ through the surface S

The flux is a measure of the, ⃗ - field lines passing through the surface . The flux
through any close surface is a measure of the total charge inside. Charges outside the
closed surface contribute nothing to the total flux.
For the point charge at the origin, the flux of ⃗ through a sphere of radius is given
by,

∫⃗ ∫ * ̂+ ̂

Therefore the flux through any surface enclosing the charge is . If instead there are a
number of charges scattered about, then the total electric field is given by the sum of the
individual fields i.e,

⃗ ∑⃗

The flux will be given by

∮⃗ ∑ ∮⃗⃗⃗⃗ ∑
 But the total charge enclosed (charge density) within the surface is

Hence, ∮⃗ which GAUSS’S LAW in Integral form.

To change this equation to the differential form, we apply the divergence theorem which
sates that;
The surface integral of a vector field over a closed surface, which is called the flux
through the surface, is equal to the volume integral of the divergence over the region
inside the surface.
∮⃗ ∫

Re –writing in terms of charge density we have,

∫

Hence, ∫ ∫ or which GAUSS’S LAW in differential form.

IMPORTANT THEOREMS
(i) DIVERGENCE THEOREM
The surface integral of a vector field V over a closed surface, which is called the flux
through the surface, is equal to the volume integral of the divergence over the region
inside the surface. i.e

∮⃗ ∫

(ii) STOKE’S THEOREM

Given a vector field, the Stoke’s theorem relates the integral of the curl of a vector V
field over some surface, to the line integral of the vector field around the boundary of the
surface .i.e

∫ ∮⃗

(iiI) FUNDAMENTAL THEOREM OF CALCULUS/GRADIENTS

States that if we can find an anti-derivative for the integrand, then we can evaluate the
definite integral by evaluating the anti-derivative at the endpoints of the interval and
subtracting. i.e

∫ or ∫

Where
DIVERGENCE AND CURL OFELECTROSTATIC FIELDS
(i) DIVERGENCE OF ELECTROSTATIC FIELDS
Consider the equation for the electric field for volume charge distribution i.e,
̂
⃗ ∫ ̂ ∫ ……................(1)

Originally the integration was over the volume occupied by the charge, but can be extended to
all space, since outside the volume. Note that -dependence is contained in ⃗ ,
thus,
Taking divergence on both sides of (1) we get,
̂
⃗ * ∫ +

̂
[ ∫ ( ) ] (since is for unprimed coordinates) …….........(2)

̂
But ( ) called the Dirac Delta function…….........(3)
and ⃗ Therefore,

…….........(4)

Therefore, ⃗ which is GAUSS’S LAW in differential form.

(ii) CURL OF ELECTROSTATIC FIELDS

Consider the equation for the electric field for volume charge distribution i.e,
̂
⃗ ∫ ̂ ∫ ……................(1)

Taking the curl on both sides of (1) we get,

̂
⃗ * ∫ +

̂
∫ (Since, is for unprimed coordinates) …….........(2)
̂
But, ( ) (Since the curl of a gradient is zero).
Hence, ⃗ .

WEEK THREE
The Electric Potential
The electric potential ⃗ at a point from a reference point O is given by,
⃗ ∫ ⃗ …………………………(1)
The potential difference between two points and is given by,

⃗ ∫ ⃗ ∫ ⃗

∫ ⃗ ∫ ⃗
∫ ⃗ …………………………(2)
From the fundamental theorem of calculus/gradients,

Hence

∫ ∫ ⃗

Since this is true for any point and then

⃗ …………………………(3)
Equation (3) is the differential form of equation (1). It states that the electric field is the gradient
of a scalar potential

Poisson's Equation and Laplace's Equation

Recall that ⃗
But ⃗ , therefore

Or

or Which is known as Poisson's Equation

When (where there’s no charge) Poisson's Equation becomes Laplace's Equation i.e

Also we know that ⃗ and or i.e the curl of a gradient is always

zero.

The Potential of a Localized Charge Distribution

Taking the reference point at infinity, the potential of a point charge at the origin is given by,
 ∫ +

In general, the potential of a point charge is

where , as always, is the distance from the charge to .

 For a collection of charges i.e discrete, the potential is,

 For a continuous distribution of charges the potential is,

 For a volume charge the potential is,

 For a surface charge the potential is,

 For a line charge the potential is,

Note that we shall be taking the reference point to be at the infinity .

Work and Energy in Electrostatics

Consider a set of stationary source charges. , as shown in figureH ow much
work is done a to move a test charge from point to point ?

Figure 8: work is done a to move a test charge

At any point along the path, the electric force on Q is = Q ⃗ . The force one must exert
in the opposite direction is -Q ⃗ The work is given by

∫ ∫ ⃗ ∫ ⃗ [ ]
Note that the work is independent of the path taken from to in such a case the
electrostatic force is said to be a "conservative force."

Rearranging the equation we have,

i.e In words, the potential difference between points a and b is equal to the work done per
unit charge required to carry a particle from a to b.
In order to bring the charge from infinity to point , the work required is,
[ ]

The Energy of a Point Charge Distribution

We wish to determine how much work is required to assemble a collection of point
charges

Figure 9: A collection of point charges

The first charge, , requires no energy, since there is no field yet to work against. To
bring in the work done is,
– where is the potential due to and is the position of .
But , hence ( )

Similarly ( ).

and ( ) and so on

Therefore the total work required is i.e

( )

The general rule is to take the product of each pair of charges, divide by their separation
distance, and sum them up i.e
 ∑∑

Where so as not to count the same pair twice. Another method is to intentionally count
each pair twice, and then divide by 2: i.e

∑∑

Another way is to express the work done by using the potential. i.e

∑ ∑

( )

∑ ⃗

∑ ⃗

Where ⃗ is the potential at point the position of due to all the other charges. That's the
work needed to assemble a configuration of point charges; it's also the amount of work needed to
dismantle the system. It represents potential energy stored in the configuration.

The Energy of a Continuous Charge Distribution

 For a volume charge density the work done is given by,

 For a line charge density the work done is given by,

 For a surface charge density the work done is given by,

From Gauss’s law, ⃗ or ⃗ Therefore, For a volume charge density the work
done is given by

∫( ⃗ )

We can make use of the product rules of vector calculus together with the appropriate
fundamental theorem. i.e
( ) ( )
Integrating over volume we have
∫ ( ) ∫ ( ) ∫

Apply the divergence theorem to the L.H.S i.e ∫ ( ) ∮( )

Therefore; ∫ ( ) ∫ ∮( )

And
[∫ ⃗ ∮ ⃗⃗⃗ ]

But ⃗ hence,

[∫ ⃗ ∮ ⃗⃗⃗ ]
This means that as the volume increases to all space, the contribution from the volume integral
goes up and that of that of the surface integral goes down to zero. Therefore;

[∫ ⃗ ]

Example
Find the energy of a uniformly charged spherical shell of total charge and radius
Solution

∫
The potential at the surface of the sphere (which is a constant) is given by,

Hence

∫ ∫
WEEK FOUR
CONDUCTORS
All materials contain charges. In an insulator, the charges are bound, whereas in a conductor
there are free (mobile) charges. In most cases these are the electrons.
A perfect conductor is a material with an unlimited number of completely free charges
Special electrostatic properties of conductors are:
 The electrostatic field inside a conductor is zero, and on the surface it is perpendicular to
the surface.

Explanation: If there was an electric field inside the conductor, that would cause the
free charges to move. As they moved, they would cause an electric field opposing the
external field. They will continue to move until the “induced field” and the external field
cancel at each point inside the conductor. On the surface, movement is restricted to the
tangential direction (along the surface), so by the same argument as before, the tangential
component of the electric field is zero i.e. the electric field can only have a component
normal (perpendicular) to the surface.

 The net charge density at any point inside the conductor is zero (and hence any net
charge resides on the surface).
Explanation: We know that, ⃗ , since ⃗ inside the conductor then .
 The conductor is an equipotential.
Explanation: Consider any two points and in or on the conductor. The potential
difference between these points, which independent of the path is,
∫ ⃗
For a path a path inside the conductor where, ⃗ , also ∫ ⃗ and therefore
conductor then , i.e equipotential
 The charge induced on the inner wall of a cavity in a conductor is equal but opposite the
charge inside the cavity.
Explanation: Recall Gauss’s law, ∮ ⃗ . Consider a Gaussian surface
completely inside the conductor but enclosing the cavity. Since, ⃗ , also ∮ ⃗
and therefore . This means that the induced charges cancel the charges within
the cavity.
 ⃗ is perpendicular to the surface, just outside a conductor otherwise, charge will
immediately flow around the surface until it destroys the tangential component.
Perpendicular to the surface, charge cannot flow, of course, since it is confined to the
conducting object.
Capacitors:
Consider an arrangement of two neutral conductors A and B. Then we transfer an amount of
charge (-Q) from A to B (+Q) as shown in figure

Figure 10: A collection of point charges

Since V is constant over a conductor i.e Both conductors will be equipotentials, the
potential difference between them is ,
∫ ⃗

Recall: ⃗ ∫ ̂

Since ⃗ is proportional to , also is is proportional to i.e

or where is the constant of proportionality called capacitance given by,

Capacitance is a geometrical quantity, determined by the sizes, shapes, and separation of the two
conductors. The SI unit is C is measured in farads (F); a farad is a coulomb-per-volt . But
this unit is inconveniently large, other practical units are the microfarad
(10- 6 F) and the picofarad (10- 12 F).
By definition, is the potential of the positive conductor less that of the negative one and is
the charge of the positive conductor hence capacitance is a positive quantity.

Example
Find the capacitance of two concentric spherical metal shells, with radii and .
Solution:
Place charge +Q on the inner sphere, and – Q on the outer one. The electric field between
the spheres is
⃗ ∫ ̂

But ⃗ ∫ ⃗ ∫ ( )

Therefore, Capacitance is

Work Needed To Charge a Capacitor:

To charge a capacitor, electrons are removed from the positive plate or surface to the negative
plate or surface thereby working against the electric field, which is pulling them back toward the
positive plate and pushing them away from the negative plate.
Suppose the charge on the positive plate is then the potential difference is,
But or

The work needed to move the next charge is . Integrating from to

we have, ∫ ∫ or

But , therefore, , where is the final potential of the capacitor.

Energy Stored In a Capacitor:

 Recall the Energy equations,
For a volume charge density the work done is given by,

 For a surface charge density the work done is given by,

∫
Now the total energy is obtained by adding contributions from the two surfaces as follows,

∫ ∫
∫ ∫

But hence

WEEK FIVE
ELECTRIC FIELDS INMATTER
Polarization
Dielectrics
 Dielectrics are insulators. All charges are attached to specific atoms or molecules-they're
tightly bound, and but they can do is move a bit within the atom or molecule.
 When placed in an electric field, ⃗ it distorts the charge distribution of a dielectric atom
or molecule it two ways: by stretching and by rotating.
Induced Dipoles
A dipole consists of two equal and opposite charges separated by a distance as
shown in figure 11
Figure 11: A dipole
Consider an atom placed in an electric field, ⃗ . Although the atom as a whole is
electrically neutral, there is a positively charged core (the nucleus) and a negatively
charged electron cloud surrounding it. These two regions of charge within the atom are
influenced by the field: the nucleus is pushed in the direction of the field, and the
electrons the opposite way. If the field is large enough, it can pull the atom apart
completely, "ionizing" it and the substance then becomes a conductor.
With smaller fields, equilibrium is soon reached, since if the center of the electron cloud
does not coincide with the nucleus, these positive and negative charges attract one
another, and this holds the atoms together.
The two opposing forces ⃗ which pulls the electrons and nucleus apart, their mutual
attraction drawing them together, reach a balance, leaving the atom polarized, with the
positive charge shifted slightly one way, and the negative the other. The atom now has a
tiny dipole moment, , which points in the same direction as ⃗ . This induced dipole
moment is approximately proportional to the field as long as the ⃗ is not too strong, thus,
⃗ or ⃗
The constant of proportionality is called atomic polarizability. Its value depends on
the structure of the atom.

Polarization
If the substance is made of neutral atoms (or non-polar molecules), the field will induce
in each a tiny dipole moment which points in the same direction as the field. If the
material is made up of polar molecules, each permanent dipole will experience a torque,
tending to line it up along the field direction and the material becomes polarized. A
convenient measure of this effect is the polarization ( ⃗ ) defined as the dipole moment
per unit volume i.e
⃗ dipole moment per unit volume i.e ⃗

For the "physical" dipole with equal and opposite charges, (±q) the dipole moment,

Where is the the vector from the negative charge to the positive one (Fig. 12).
 Figure 12: A physical dipole
The dipole moment of a collection of discrete point charges is given by,

The dipole moment for continuously distributed charges is given by,

∫

Example
Calculate the atomic polarizability of a model atom consisting of a point nucleus (+q)
surrounded by a uniformly charged spherical cloud (-q) of radius a.

Solution
In the presence of an external field, ⃗ , the nucleus will be shifted slightly to the right
and the electron cloud to the left, as shown in Figure 12 (b).

Figure 12: A model atom

The ⃗ field at a distance d from the center of a uniformly charged sphere is

But =( ) ⃗ and ⃗ therefore, by comparison,

The Field of a Polarized Object
Bound Charges

Consider a polarized material with Polarization, ⃗ . We want to find the field produced by this
material due to polarization. It's easier to work with the potential. For a single dipole we
have equation
̂
⃗

But ⃗ or ⃗ Integrating we get, ⃗ . Therefore,

⃗ ̂ ̂ ̂
⃗ ∫ , But ( ) . Hence,

̂
⃗ ∫⃗ ( ) Integrating by parts using the product rule and applying the
divergence theorem, i.e

∫ ( ) ∫ ( ) ∫ ∮( )
We get,

⃗ ∮ ⃗ ∫ ⃗

 The first term looks like the potential of a surface charge i.e.

⃗ ⃗ ̂ which is the potential of surface charge

 The second term looks like the potential of a volume charge i.e.

⃗ ⃗ which is the potential of volume charge

Therefore we can write the potential as,

⃗ ∮ ∫

This means that the potential (and hence the field) of a polarized object is the same as that
produced by a volume charge density ⃗ plus a surface charge density ⃗ ̂
Where
⃗ are bound charges, accumulated in the in the volume and
⃗ ̂ are bound charges, accumulated in the on the surface
The Electric Displacement
Within the dielectric, then, the total charge density can be written:

Where is the total bound charge, while are free charges

From Gauss’s law, ⃗ or ⃗ hence,

or ⃗

But ⃗ therefore,

⃗ ⃗ or ⃗ ⃗ which becomes,

( ⃗ ⃗)

Or ⃗⃗ which is Gauss’s law in differential form in terms of ⃗⃗

where ⃗⃗ ⃗ ⃗ is called the electric displacement.

We can also write ∮ ⃗⃗ the total free charge enclosed in a volume. This equation
gives the Gauss’s law in integral form in terms of ⃗⃗ .

Linear Dielectrics
For many substances, provided ⃗ is not too strong, the polarization is proportional to the field,
i.e,

⃗ ⃗ where is the electric susceptibility of the medium and it is dimensionless. The

value of depends on the microscopic structure of the substance and also on external
conditions such as temperature.
Materials that obey the equation, ⃗ ⃗ are called Linear Dielectrics.
Recall; ⃗⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗
Where is called the permittivity of the material.
In vacuum, where there is no matter to polarize, the susceptibility is zero, and the permittivity is
. That's why is called the permittivity of free space and =8.85 x 10-12 C 2 /N.m2 .
Dividing through by we get
which is called the relative permittivity, or dielectric constant, of the

material
WEEK SIX
Magnetic fields
The Lorentz force law
For charges in motion, experiments have shown that the force between them is not given by
Coulomb’s law. The magnetic force on a charge moving with velocity in a magnetic field ⃗
is given by ⃗ ⃗ ⃗⃗ which is called Lorentz force law. The Lorentz force ⃗ ⃗ ⃗⃗ , is

the magnetic analogue of the electric law ⃗ ⃗⃗ .

In the presence of both electric and magnetic fields the net force on the charge would be,

⃗ ⃗ ⃗⃗ ⃗ (⃗ ⃗⃗ ) ⃗

Work done by Magnetic forces

Magnetic forces do no work.
Proof.
If a charge moves a distance , then the work done is

∫ ⃗

Since ⃗ is perpendicular to the velocity of the charge, then

( ⃗) | ⃗ || |
Magnetic forces may alter the direction in which a particle moves, but they cannot speed it
up or slow it down. This means that although magnetic forces are able to change the velocity
of a charge, they are unable to change the kinetic energy and hence speed of a charge
 If the charge is not moving ( ) there is no magnetic force, even if the magnetic field is
not zero!

Currents
 The current in a wire is the charge per unit time passing a given point.
 Current is measured in coulombs-per-second, or amperes (A): ⁄
(i) Line Current Density
 For a line charge traveling down a wire at speed , the current is given by,

 The magnetic force on a segment of current-carrying wire is given by,

∫ ⃗ =∫ ⃗ ∫ ⃗⃗⃗⃗ ⃗ ∫ ⃗ ∫ ⃗
 Figure 13: Line Currents Density

(ii) Surface Currents Density

 Denoted by ⃗ charge traveling down a wire at speed , the current is given by,
 ⃗ is the current per unit width-perpendicular-to-flow. ⃗

 For mobile surface charge density with velocity then ⃗ ⃗

Figure 14: Surface Current Density

 Therefore

∫ ⃗ ∫ ⃗ ∫ ⃗ ⃗

(iii) Volume currents Density

 Denoted by charge traveling down a wire at speed , the current is given by,
 is the current per unit area-perpendicular-to-flow.
 Figure 15: Volume Currents Density
Conservation of charge and the continuity equation
 For mobile volume charge density with velocity then ⃗

∫ ⃗ ∫ ⃗ ∫ ⃗

Using or ∫ or by divergence theorem,

∮ ∫ ∫ ∫( )
The minus sign means that outward flow decreases the charge left inside the volume V.
Therefore, this is called the continuity equation, a precise mathematical statement

of local conservation of charge.

The Biot-Savart Law

Steady Currents
Stationary charges produce electric fields that are constant in time; hence the term electrostatics.
Steady currents produce magnetic fields that are constant in time; the theory of steady currents is
called magnetostatics.
Steady current mean a continuous flow that has been going on forever, without change and
without charge accumulating anywhere along the conductor.
For steady currents,

(Electrostatics)

(Magnetostatics)

Note that a moving point charge cannot possibly constitute a steady current. A moving point
charge does not produce a static field.
When a steady current flows in a wire, its magnitude must be the same all along the
line; otherwise, charge would be accumulating somewhere, and it wouldn't be a steady current.
Therefore I is the same in magnetostatics, and hence the continuity equation becomes,
The Magnetic Field of a Steady Current
The magnetic field ⃗ of a steady line current is given by the Biot-Savart law i.e
̂ ̂
⃗ ∫ ∫

Figure 16: Line Currents Density

Where the integral is along the path of the current ( , is the element of length, ̂ is
the vector from the source to point . The constant is called the permeability of free
space : ⁄ . The unit of ⃗ is Tesla (T) and 1T=1N/Am.

Note that, ̂ | || ̂|

Example
Find the magnetic field at O due to the loop of shown below which carries a steady
current .

Solution:
0 i  r   dl   Rˆ 0 i dl   Rˆ
B r      (because the current is constant)
4 C R2 4 C R 2
 0 i  dl   Rˆ dl   Rˆ dl   Rˆ dl   Rˆ 
B r           ẑ into the page
4 C1 R 2 C2 R2 C3 R2 C4 R2 

0 i  0 dl zˆ 0 dl    zˆ   0 i  dl  dl  
B r    2   2   2    zˆ   2   2 
4 C1 R C2 a C3 R C4 b 2
 4 C2 a C4 b 

0 i  1 1   i  1  a 1  b  0 i  1 1 
B r   zˆ  2  dl   2  dl   0 zˆ  2    zˆ
4  a C2 b C4  4  a 2 b2 2  8  a b 

Note:
 For surface currents the Biot-Savart law becomes
⃗ ̂
⃗ ∫

 For volume currents the Biot-Savart law becomes

̂
⃗ ∫

 The superposition principle applies to magnetic fields just as it does to electric fields, for
a collection of source currents; the net field is the (vector) sum of the fields due to each
of them taken separately.

The Divergence and Curl of magnetic field ⃗⃗

(i) The Divergence of ⃗⃗

Consider the Biot-Savart law for a volume current.
̂
⃗ ∫

Applying the divergence on both sides

̂
⃗ * ∫ +

̂
⃗ *∫ ( ) +……the integral is over the primed coordinates while the

divergence is over the unprimed coordinates.

 Applying the product rule, (⃗ ⃗) ⃗ ( ⃗) ⃗ ( ⃗ ) we have,

̂ ̂ ̂
( ) ( ) ( )

̂
But because does not depend on the unprimed variables. Also
̂
Which means, ( ) . Therefore, ⃗ i.e the divergence of magnetic field

is zero.
Implications of   B  0 :
Theorem

Divergence-less (solenoidal fields): The following are equivalent:

(a)   F  0  A  0
(b) There is some vector function ( A ) of which F is the curl i.e. F    A

(c)  F  da is independent of the surface, for some fixed boundary line

S

(d)  F  da  0 for any closed surface

Since (a) is true, we have B    A and all the rest are also true.

0 J  r    Rˆ
 We call A the magnetic vector potential. Now B  r    d 
4 V  R2
 J  r   1  1  Rˆ  J  r    Rˆ
But   
 R  R    
  J  r   J  r       0  J  r     2  
 R R2
   R 

0 J  r    Rˆ 0  J  r    0 J  r   
So B  r    d         d        d   
4 V  R2 4 V   R   4 V  R
   

0 J  r  
we see that Ar    d 
4 V  R

 We can also easily show that (a) implies (d):

 B  da     B d where V is the volume inside the closed surface, by the divergence
V

theorem.

But using (a),   B  0 and so  B  da   0 d  0

V

The magnetic flux through any closed surface is zero:  B  da  0

(ii) The Curl of ⃗⃗

Consider the Biot-Savart law for a volume current.
̂
⃗ ∫

Applying the curl on both sides we have,

̂ ̂
⃗ * ∫ + ∫ ( )

Applying the product rule, (⃗ ⃗) (⃗ )⃗ (⃗ )⃗ ⃗( ⃗ ) ⃗( ⃗)

̂ ̂ ̂ ̂ ̂
( ) ( ) ( ) ( ) ( )

We drop the terms involving the derivatives of , since does not depend on .

Hence
̂ ̂ ̂ ̂ ̂
( ) ( ) ( ) ( ) ( )

̂ ̂
But ̂ and ( )

Therefore

̂
⃗ ∫ ( )

̂ ̂
∫* ( ) ( ) +
 ∫ ⃗ ⃗

∫ ⃗ ⃗

∫ (⃗ ⃗ ) ∫ (⃗ ⃗ )

Therefore,
⃗

Ampere's Law
The equation for the curl of B, i.e ⃗ is called Ampere's Law (in differential
form). It can be converted to integral form by applying Stokes' theorem,

∫( ⃗ ) ∮⃗ ∫⃗

But ⃗ which is the total current passing through the surface enclosed by the Amperian
loop. Therefore
∮⃗ which is Ampere's Law (in integral form)

Implications of   B  0 J :

We can integrate both sides over the same surface:

S
 
   B  da   0 J  da  0  J  da  0 ithrough S .
S S


But using Stokes theorem,    B  da   B  dl
S


Ampere’s Law  B  dl  0 ithrough S

Notice that Ampere’s law tells us that by computing  B  dl just around the circumference of a
surface, we can find out how much current flows through the surface!

 Ampere’s Law is to magnetostatics, what Gauss’s law is to electrostatics

 Qinside V
 B  dl  0 ithrough S  E  dl 
0

 It is often used to find magnetic fields for symmetrical current distributions

 But beware, the symmetry can be much less obvious
 The symmetry is for physical effects i.e. forces

 The magnetic B -field is always perpendicular to the magnetic force, since F  Qv  B

WEEK SEVEN
WEEK EIGHT